Problem: We are given that $\dfrac{dy}{dx}=\dfrac{\text{sin}(y)}{x}$. Find an expression for $\dfrac{d^2y}{dx^2}$ in terms of $x$ and $y$. $\dfrac{d^2y}{dx^2}=$
Notice that the equation defines $y$ implicitly—we don't have an explicit expression for $y$ in terms of $x$. So we will have to use implicit differentiation. If we differentiate the equation once, we will have $\dfrac{d^2y}{dx^2}$ on the left-hand side of the equation. This is what we get after we differentiate each side of the equation: $\dfrac{d^2y}{dx^2}=\dfrac{x\cdot\text{cos}(y)\dfrac{dy}{dx}-\text{sin}(y)}{x^2}$ The expression we got for $\dfrac{d^2y}{dx^2}$ isn't in terms of only $x$ and $y$ because it includes $\dfrac{dy}{dx}$ in it. Let's plug the original equation ${\dfrac{dy}{dx}=\dfrac{\text{sin}(y)}{x}}$ in the new equation to obtain an expression in terms of only $x$ and $y$ : $\begin{aligned} \dfrac{d^2y}{dx^2}&=\dfrac{x\cdot\text{cos}(y){\dfrac{dy}{dx}}-\text{sin}(y)}{x^2} \\\\ &=\dfrac{x\cdot\text{cos}(y)\cdot{\dfrac{\text{sin}(y)}{x}}-\text{sin}(y)}{x^2} \\\\ &=\dfrac{\text{cos}(y)\text{sin}(y)-\text{sin}(y)}{x^2} \end{aligned}$ In conclusion, this is an expression for $\dfrac{d^2y}{dx^2}$ in terms of $x$ and $y$ : $\dfrac{d^2y}{dx^2}=\dfrac{\text{cos}(y)\text{sin}(y)-\text{sin}(y)}{x^2}$